A ball of mass 2,kg and another of mass 4,kg | vedclass

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(C) The kinetic energy $(KE)$ of an object is given by the formula $KE = \frac{1}{2}mv^2$.Since both balls are dropped from rest and fall through the

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$1:2$

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(C) The kinetic energy $(KE)$ of an object is given by the formula $KE = \frac{1}{2}mv^2$.Since both balls are dropped from rest and fall through the same height $(h = 30\,ft)$,they will have the same final velocity $(v)$ according to the equation of motion $v^2 = u^2 + 2gh$,where $u = 0$.Since $v$ is the same for both balls,the kinetic energy is directly proportional to the mass $(KE \propto m)$.Therefore,the ratio of their kinetic energies is $\frac{KE_1}{KE_2} = \frac{m_1}{m_2}$.Given $m_1 = 2\,kg$ and $m_2 = 4\,kg$,the ratio is $\frac{2}{4} = \frac{1}{2}$.

$A$ ball of mass $2\,kg$ and another of mass $4\,kg$ are dropped together from a $60\,ft$ tall building. After a fall of $30\,ft$ each towards the earth,their respective kinetic energies will be in the ratio of:

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